If →a=^i+^j+^k,→b=^i+^j,→c=^i and (→a×→b)×→c=λ→a+μ→b, then λ+μ=
A
1
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B
0
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C
−1
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D
2
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Solution
The correct option is A0 →a×→b=−^i+^j →c=^i Hence (→a×→b)×→c =(−^i+^j)×(^i) =−^k λ→a+μ→b =(λ+μ)^i+(λ+μ)^j+λ^k Comparing coefficients, we get λ+μ=0 and λ=−1 Hence μ=1